tìm các số nguyên x y biết: xy+2y+x=1
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Lời giải:
$xy-x+2y=2$
$\Rightarrow (xy-x)+(2y-2)=0$
$\Rightarrow x(y-1)+2(y-1)=0$
$\Rightarrow (x+2)(y-1)=0$
$\Rightarrow x+2=0$ hoặc $y-1=0$
$\Rightarrow x=-2$ hoặc $y=1$
a.
xy + 3x - 2y - 6 = 5
=>x(y + 3) - 2(y + 3) = 5
=>(x - 2)(y + 3) = 5.
Vì x, y thuộc Z nên x - 2, y + 3 thuộc Z
=> x - 2, y + 3 thuộc ước nguyên của 5
Lập bảng :
x - 2 | -5 | -1 | 1 | 5 |
y + 3 | -1 | -5 | 5 | 1 |
x | -3 | 1 | 3 | 7 |
y | -4 | -8 | 2 | -2 |
Vậy ......
b. Làm tương tự câu a.
c. Ta có x + y = 3 và x - y = 15
Bài này là tổng hiệu của cấp 1, áp dụng cách làm đó thì ta được số lớn là x = (3 + 15) : 2 = 9
Số bé là y = 9 - 15 = -6
d. Ta có : |x| + |y| = 1
=>|x| = 1 - |y|
Vì |x|, |y| >= 0 và |x| = 1 - |y| nên 0 =< |x|, |y| =< 1
Vì x, y thuộc Z nên x = 0 thì y = 1 hoặc -1 và ngược lại y = 0 thì x = 1 hoặc -1
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a) pt <=> (2x-1)(2y+3)=7
TH1: 2x-1=7 và 2y+3=1
<=> x = 4 và y = -1
TH2: 2x - 1 = -7 và 2y + 3 = -1
<=> x = -3 và y = -2
TH3: 2x-1=1 và 2y+3=7
<=> x = 1 và y=2
TH4: 2x-1=-1 và 2y+3=-7
<=> x=0 và y=-5
câu a thì em nhân 2 vế rồi loại trừ nhé còn câu b đắt nhân tử chung là y ra ngoài xong đổi vế 6 sang trái là -6 xong đến đấy chắc biết rồi ^^
a) (x-3) (2y+1)=10
=> x-3 và 2y+1 thuộc Ư(10)={1;2;5;10;-1;-2;-5;-10}
Bn tự kẻ bảng rồi làm nha
b) xy+2y=6
y(x+2)
=>y và x+2 thuộc Ư(6)={1;2;3;6;-1;-2;-3;-6}
Đến đây bn tự làm nha
a)
x | 1 | -1 | 12 | -12 | 2 | -2 | 6 | -6 | 3 | -3 | 4 | -4 |
y-3 | -12 | 12 | -1 | 1 | -6 | 6 | -2 | 2 | -4 | 4 | -3 | 3 |
y | -9 | 15 | 2 | 4 | -3 | 9 | 1 | 5 | -1 | 7 | 0 | 6 |
b)
x | 1 | -1 | 3 | -3 | 7 | -7 | 21 | -21 |
y | -21 | 21 | -7 | 7 | -3 | 3 | -1 | 1 |
c)
2x-1 | 1 | -1 | 5 | -5 | 7 | -7 | 35 | -35 |
2y+1 | -35 | 35 | -7 | 7 | -5 | 5 | -1 | 1 |
x | 1 | 0 | 3 | -2 | 4 | -3 | 18 | -17 |
y | -18 | 17 | -4 | 3 | -3 | 2 | -1 | 0 |
e)
2x+1 | 1 | -1 | 5 | -5 | 11 | -11 | 55 | -55 |
3y-2 | -55 | 55 | -11 | 11 | -5 | 5 | -1 | 1 |
x | 0 | -1 | 2 | -3 | 5 | -6 | 27 | -28 |
y | loại | 19 | -3 | loại | -1 | loại | loại | 1 |
Những câu còn lại mk hổng bt làm đâu
a) (x-3).(2y+1)=10
=>x-3 và 2y+1 thuộc Ư(10)={1;2;5;10;-1;-2;-5;-10}
Bn tự kẻ bảng làm nha
b) xy+2y=6
y(x+2)=6
=>y và x+2 thuộc Ư(6)={1;2;3;6;-1;-2;-6;-3}
Bn tự kẻ bảng làm nốt nha
Nếu mk làm sai thì xl bn nhìu